Assume the acceleration of the object is a(t) = −9.8 meters per second per second. (Neglect air resistance.)

Question

A baseball is thrown upward from a height of 3 meters with an initial velocity of 8 meters per second. Determine its maximum height. (Round your answer to two decimal places.)

Answer 1

a(t) = -9.8

acceleration is the derivative of velocity which is the derivative of distance

a(t) = v'(t) = h"(t)

v(t) = h'(t) = integral of a(t) = -9.8t + c where c = unknown constant variable.

then integrate again

integral of velocity = integral of -9.8t = -4.9t^2 + ct + k = h(t) c= initial velocity, k = initial height

h(t) = -4.9t^2 + 8t + 3

h'(t) = v(t) = -9.8t + 8 =0

t= 8/9.8= 4/4.9 seconds when height is the maximum

max h = h(t) = -4.9t^2 + 8r + 3 = -4.9(4/4.9)^2 + 8(4/4.9) + 3 = -3.26 + 6.53 +3 = 6.27 meters

maximum height = 6.27 meters